so sanh a va b: A= (3+1)(3`2+1)(3`4+1)(3`8+1)(3`16+1) B=3`32-1 04/10/2021 Bởi Iris so sanh a va b: A= (3+1)(3`2+1)(3`4+1)(3`8+1)(3`16+1) B=3`32-1
Ta có: ` A=(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) ` ` <=> 2A=2(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) ` ` <=> 2A=(3-1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) ` ` <=> 2A=(3^2-1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) ` ` <=> 2A=(3^4-1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) ` ` <=> 2A=(3^8-1)(3^8 + 1)(3^{16} + 1) ` ` <=> 2A=(3^{16}-1)(3^{16}+ 1) ` ` <=> 2A=3^{32} – 1 ` Mà ` B = 3^{32} – 1 ` ` => 2A = B ` ` => A<B ` Bình luận
Giải thích các bước giải: A=(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=2(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3-1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3^2-1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3^4-1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3^8-1)(3^8 + 1)(3^16 + 1) 2A=(3^16-1)(3^16 + 1) 2A=3^32 – 1 = B => A<B ok nhé Bình luận
Ta có:
` A=(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) `
` <=> 2A=2(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) `
` <=> 2A=(3-1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) `
` <=> 2A=(3^2-1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) `
` <=> 2A=(3^4-1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) `
` <=> 2A=(3^8-1)(3^8 + 1)(3^{16} + 1) `
` <=> 2A=(3^{16}-1)(3^{16}+ 1) `
` <=> 2A=3^{32} – 1 `
Mà ` B = 3^{32} – 1 `
` => 2A = B `
` => A<B `
Giải thích các bước giải:
A=(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=2(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3-1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3^2-1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3^4-1)(3^4 + 1)(3^8 + 1)(3^16 + 1) 2A=(3^8-1)(3^8 + 1)(3^16 + 1) 2A=(3^16-1)(3^16 + 1) 2A=3^32 – 1 = B => A<B
ok nhé